`BaCl_2+H_2SO_4->BaSO_4+2HCl`
`n_{BaCl_2}=(10,4%.300)/(137+35,5.2)=0,15mol`
`n_{H_2SO_4}=(200.9,8%)/98=0,2mol`
`n_{BaCl_2}<n_{H_2SO_4}`
`n_{BaSO_4}=n_{BaCl_2}=0,15mol`
`m_{BaSO_4}=0,15.(137+32+16.4)=34,95g`
`mdd_{spu}=300+200-34,95=465,05g`
`n_{H_2SO_4du}=0,2-0,15=0,05mol`
`%m_{H_2SO_4du}=(0,05.98)/(465,05%)=1,05%`